6:40 AM
Hi All...
Hello @JohnRennie Sir

@123 Hi :-)

@JohnRennie Nice to see you sir...

And you :-)

I have a question... confusion to tell my students.
also my own questions.

OK ... ?

6:42 AM
Moment/Torque is depend on reference point let say O , perpendicular distance to the line of action of force.
1) This reference point can be inside the body
2) Outside the body
3) and can be fixed in both above cases
My confusion is that what is the benefits of having choosing freely reference point in torque formula.
Because when we change reference point for the same rotating object it gives us different value every time. It does not have unique answer as in force.

All the angular properties like torque and angular acceleration can be computed about any point, but some point are more useful than others.

@JohnRennie Great i need these answer which clear the topic. Thank you

The angular version of Newton's second law τ = Iα is obeyed no matter where you take your reference point, but in most cases taking the point anywhere except for the centre of rotation is not very useful.

Is it also possible we fixed a rigid body at some reference axis within the body/outside the body and body is rotating about its reference axis. and we calculate torque about any other reference point.

If you calculate your torques, angular accelerations, etc about the axis of rotation then everything works out nice and simply, so of course this is what we tell students to do. You can use any axis you want, but then the calculation will get messy.

6:50 AM
@JohnRennie Great , these answer help me to read the book in more intuitive and understandable way. here CENTER OF ROTATION means reference axis at which body is rotating it could be fixed/free?

@123 yes

But i read angular acceleration, rotational inertia are tensor quantities, Why these are tensor quantities?

Is angular acceleration a tensor? I would have thought not. I'll have to Google it.

Rotational Inertia is tensor.. I don't know angular acceleration is tensor or not.
I thought because rotational inertia is tensor the angular acceleration should also be tensor because $\tau = I \alpha$

The moment of inertia is indeed a tensor. But as far as I know angular velocity and angular acceleration are vectors, or more precisely pseudovectors though as far as students are concerned we can just take them to be vectors.

6:58 AM
@JohnRennie Okay... It means in torque formula. tensor (rotational inertia) is multiplying with pseudovector (angular acceleration). The output torque is vector.

Yes. In general a rank 2 tensor is something that relates two vectors.
In this case we have τ = I α where α and τ are vectors, and I is the rank 2 tensor that relates them.
You can easily see that I cannot be a simple number because the moment of inertia changes with the axis i.e. it is different in different directions.

@JohnRennie Great...

But this isn't something students need to worry about until they get to university. Although tensors are basically simple they can be confusing when you first study them. I would not mention tensors to your students unless you want to confuse them!

Pls tell me in solving physical problem, is it important to select carefully right choice of reference system (rectangular, polar, circular, cylindrical etc..) also their right choice of choosing a position of origin and set a orientation of reference system.
I read it in klepenner & kolenkow. If we carefully meet the above condition it will simplify our mathematics and solution.

Yes, in general choosing the right way to formulate a problem will make it simpler. Knowing the right way is something you can only build up with experience.

7:07 AM
@JohnRennie You are right. I think tensor as matrix , how matrix behave when it multiply with vector or another matrix. Is this the right way of thinking tensors? Because it give me intuition of tensor and behavior of tensor on other quantities.

Tensors and vectors are frequently written as matrices, and this is a nice way to represent them because then we just use the usual rules for matrix algebra.

@JohnRennie Yes but i only saw vectors as 1xn or nx1 matrix, and tensors as nxn matrix. Is this true in general?

@123 Yes

or vectors can be more than 1xn or nx1

Vectors are written as n x 1 matrices
Then in matrix form an equation like τ = I α becomes:
(n x 1) = (n x n) (n x 1)

7:13 AM
@JohnRennie Thanks science and math can be understand intuitively. I have heavily disagreement. Why teachers don't teach these subjects like that. We need teachers like you and curius mind
Is there only two types of motion 1) Translation 2) Rotation
@JohnRennie Why books add 3) Oscillatory Motion as third type

Do they?

@JohnRennie Yes in our 10th and 12 standard books they wrote 3 types of motion.
How can we understand oscillatory/ vibratory motion in term of translation or rotation? Under which type of motion it fall.

7:34 AM
How does the textbook present them?

@user178758 Textbook present oscillatory motion as third type of type.

Does the book compare and contrast the similarities and differences?

@user178758 No.. If there are only two types of motion. Oscillation must fall any one of them or both.
In physics, motion is the phenomenon in which an object changes its position. Motion is mathematically described in terms of displacement, distance, velocity, acceleration, speed, and time. The motion of a body is observed by attaching a frame of reference to an observer and measuring the change in position of the body relative to that frame with change in time. The branch of physics describing the motion of objects without reference to its cause is kinematics; the branch studying forces and their effect on motion is dynamics. If an object is not changing relatively to a given frame of reference...
In this type wiki article they oscillation as separate type.
Also circular motion as separate type and projectile motion also

that's the standard way of presenting them

@user178758 Are you asking or telling me?

7:44 AM
I am stating a fact :-)

@user178758 Ooookay.. We were read in 10th standard only three type of motion.

it will help them to learn

Okay

2 hours later…
9:28 AM
Why constant "k" is used in newton's second law of motion F = k ma . What is the purpose of it. When constant is needed physically or mathematically in this case or generally.

@123 Where did you see this? I have never seen Newton's second law written like this

@B.Brekke I have seen in many books also teachers explain it in different ways . I don't understand who is correct.
Principle of mechanics synge in chapter 1. also in physics book.

9:44 AM
Are you confusing it with Hooke's law?

some teachers said we create two new physical quantities here FORCE and MASS that's why it is needed. POM synge book said by selection proper unit system we can set k = 1. book relate it to unit system.

@123 Okay, in that case the $k$ constant is used to relate the unit of force with the unit of mass. If you use Newtons for force and kilograms for mass, as we often do, you set $k=1$. However, if you use Newtons for force and grams for mass, you need a conversion factor $k$.
I prefer to express my variables in the SI system, and then use Newton's second law. In that way, I can always use $k=1$.
Note that the units of acceleration also matters

10:09 AM
@123 we don't "create" physical quantities; we measure physical quantities with standardized units, right?

10:27 AM
2

10:46 AM

3 hours later…
1:44 PM
Do non rel. spinors transform under $SL(2,\mathbb{C})$?

1:55 PM
SU(2) afaik

2:06 PM
Okay so if I were to construct an action with a potential term that transforms as a scalar under that group, would I just have $\epsilon_{\alpha \beta}\psi^\alpha \psi^\beta$?
There must be a Hermitian adjoint there somewhere right?

2:58 PM
While defining operator products of energy momentum tensor and fields in OPE, we take them to be radially ordered(time ordered). Becker,Becker,Schwarz book on string theory says that this is done so that the product have convergent series expansions...Why would time ordering imply convergent series expansion?

3:15 PM
Maybe it can't be what I suggested because it isn't real

3:31 PM
What about $\epsilon_{\alpha \beta} \psi^\alpha \overline{\psi}^\beta$ where $\overline{\psi}^\beta = (\epsilon \psi^\dagger)^\beta$?
@ACuriousMind What do you think?

4:00 PM
Does anyone know what to do?

1 hour later…
5:23 PM
Sorry, I forgot to mention that $\psi^\alpha$ are non relativistic quantum fields

5:57 PM
I don't know. What I wrote doesn't seem to be invariant under $SU(2)$
But that would mean that $\psi_1 \psi_1^* + \psi_2 \psi_2^*$ isn't which would mean that I've made a mistake I think

2 hours later…
7:29 PM
@DIRAC1930 just use the normal spinor terms with the $\gamma$s and everything, you just have that the $\gamma$ are simply Pauli matrices for 3d

How exactly do we construct invariant terms? We have the Hermitian inner product from QM but also the transformation of spinor fields classically
We want a term whose expectation value is invariant too?
right

I'm not sure what you mean, exactly
if the operator is invariant (classically), then the expectation value is, too
the representations on the fields and on the space of states need to be compatible, otherwise QFT doesn't work

7:45 PM
Don't quantum fields transform like $U \Psi U^\dagger$ or something though?
Oh but that does something like $U \Psi U^\dagger = D \Psi(\dots)$
where $D$ is a representation of the group
right?

yes, that's one of the Wightman axioms and what I mean by "compatible"

So for $SU(2)$ it is correct to say that $\epsilon_{\alpha \beta}\psi^\alpha \psi^\beta$ is an invariant
But where does $\psi^\dagger$ come in

it's invariant, that doesn't mean it's real
you want your Lagrangian to be real

But if I write $\epsilon_{\alpha \beta} \psi^\alpha {\psi^\dagger}{}^\beta$ it's not invariant correct?
I need the $\psi^\dagger{}^\beta$ to be something else

you probably shouldn't write the index $\beta$ upper on the $\dagger$

7:50 PM
Okay sorry

because conjugating changes how it transforms

$\epsilon_{\alpha \beta}\psi^\alpha \psi^*{}^\beta$
Does that make sense?

where, exactly, is the difference between $\ast$ and $\dagger$ here?

Well dagger comes in when you write the above as $\psi \epsilon \psi^\dagger$ in matrix notation
$*$ is just complex conjugate

sure, but we aren't in matrix notation
you wrote down components

7:52 PM
Yes that was my mistake

what is the difference between the components of $\psi^\ast$ and $\psi^\dagger$ that you think makes this valid where it wasn't with $\dagger$?
my point about the index position was that conjugating the vector changes the rep from the "normal" to the conjugate one
and usually in this notation, the upper/lower distinction should be precisely that between a rep and its conjugate
otherwise you'll have to explain to me what the upper/lower mean in your notation

So does the thing transform like $\epsilon_{\alpha \beta} (U \psi)^\alpha (U \psi)^*{}^\beta$?

why is the $\epsilon$ not transforming at all?

I don't know what I'm doing thats why lol
I don't know how to transform it

well, but why are you writing the indices on it like that/why are you trying to construct an invariant with it?

7:58 PM
Because I want to see what the potential term properly looks like in non-rel QFT

I'm not really sure why you're trying to use an $\epsilon$ at all, really
what "potential term"?

$\epsilon$ is the invariant tensor of $SU(2)$
It's used to raise and lower indicies
In the Lagrangian

I mean, your $SU(2)$ representation has an inner product (it's unitary), no?
just take the inner product of $\psi$ with itself
that's invariant and it's a $\psi^2$ "potential term"
square this to get arbitrary even terms
this is why I don't like index notation :P

So the only inner product I should care about is the one invented by Dirac or whoever?
The Hermitian inner product of QM

no, the representation on the target space of fields isn't "quantum"
it's of course isomorphic to representations you might use in QM, but we're talking about the classical field here
if you can't construct a classically useful Lagrangian you can't even begin to do QFT, after all

8:05 PM
In non rel QFT, classical fields aren't really mentioned at all hence my confusions
So the way to do things is canonical quantization there too?

Weinberg would disagree, but sure :P

But $\epsilon_{\alpha \beta} \psi^\alpha \psi^\beta$ is invariant but it's not the correct one used

again, it might be invariant, but I'm pretty sure it isn't real

Before I move on to making it real, $\epsilon$ doesn't transform right?

I don't understand how it's supposed to not transform and yet has two free indices
what do the indices mean if it doesn't transform?

8:13 PM
I thought that it just does the contraction

it might be such that it is the same after transformation (just like the Minkowski metric is invariant under Lorentz transformations), but that's different from not transforming at all
yeah, it's like a metric
but a metric is not an invariant as such - it's not a scalar

My confusion is that when finding the invariant tensor, we assume that it transforms into itself

yeah, think of it exactly like the Minkowski metric in SR

So if I had a random tensor of numbers $A_{\alpha \beta}$, it's impossible to say how it transforms?

???
if you write down the indices, that tells you how it transforms!"
the whole point of upper/lower index notation is that the indices tell you how many transformations/inverse transformations to add if you do a transformation

8:17 PM
$U_{\alpha}{}^n U_{\gamma}{}^m A_{nm}$
SO it transforms like that

yes, exactly
just because that's equal to the original $\epsilon$ in the case of the specific $\epsilon$ doesn't mean it doesn't transform
and that's what "eats" the two transformations from the two $\psi$ and would make your term invariant

Okay thanks
Now all we need to do is find something that a) transforms as a scalar under $SU(2)$, and b) that is real

8:35 PM
Ah I can't do it
But if I could, would I then just replace the fields with field operators and be done?

2 hours later…
10:53 PM
The classical example of an SU(2) scalar is the Higgs field
Although I guess that one is a complex field

11:24 PM
I'm a bit confused by the inner product
$\epsilon_{\alpha \beta} \psi^\alpha \psi^\beta = \psi^\alpha \psi_\alpha$
However
$\epsilon_{\alpha \beta} \psi^\alpha \psi^\beta = -\epsilon_{ \beta \alpha} \psi^\alpha \psi^\beta = -\psi_\beta \psi^\beta$
because $\epsilon = -\epsilon^T$. I don't get why they aren't the same

what do we mean when we say a point moving in phase space?

11:58 PM
Do I have to take into account that spinor fields are anticommuting?